Diborane (B2H6) has been considered as a possible rocket fuel. Calculate DH° for the reaction B2H6(g) ---> 2B(s) + 3H2(g) using the following data: DH° (kJ) 2B(s) + 1.5O2(g) --> B2O3(s) –1273 B2H6(g) + 3O2(g) --> B2O3(s) + 3H2O(g) –2035 H2(g) + O2(g) --> H2O(g) –242 A. DH° = –36 kJ B. DH° = +36 kJ C. DH° = –86 kJ D. DH° = –86 kJ

A. DH° = –36 kJ

Explanation:

It is possible to obtain DH° of a reaction by the sum of DH° of half reactions. The DH° of the reaction:

B₂H₆(g) → 2B(s) + 3H₂(g)

Could be obtained from:

(1) 2B(s) + 1.5O₂(g) → B₂O₃(s) DH° = –1273kJ

(2) B₂H₆(g) + 3O₂(g) → B₂O₃(s) + 3H₂O(g) DH° = –2035kJ

(3) H₂(g) + 0.5O₂(g) → H₂O(g) DH° = –242kJ

The sum of (2) - (1) gives:

B₂H₆(g) + 1.5O₂(g) → 2B(s) + 3H₂O(g) DH° = -2035kJ - (-1273kJ) = -762kJ

Now, this reaction - 3×(3):

B₂H₆(g) → 2B(s) + 3H₂(g) DH° = -762kJ - (3×-242kJ) = -36kJ

Thus, right answer is:

A. DH° = –36 kJ

i need mutiple choices

d

explanation: